0

I wish to prove that $(x-2)^3 = 3 \text{ mod 13}$ admits no solutions. Admittedly, I could always plug in the numbers $0, 1, 2, ..., 12$ and show that for each of them $(x-2)^3$ is not congruent to $3$, but I would wish for a cleverer method than this...

I have thought of showing (if this indeed sufficient) that $3 + 13k$ can never be written as a perfect cube.

J. W. Tanner
  • 60,406
V. Elizabeth
  • 446
  • your second paragraph works. Just find the cubes $\pmod {13}.$ Like this: $1^3 = 1.$ $ 2^3 = 8. $ and so on – Will Jagy Sep 19 '23 at 03:35
  • @WillJagy Is the best method to show that the only possible residues of cubes modulo 13 are 0, 1, 8 by checking every number between 0 and 12? – V. Elizabeth Sep 19 '23 at 03:38
  • 1
    @V.Elizabeth: you missed some cubes, e.g., $4^3$ and $7^3$ – J. W. Tanner Sep 19 '23 at 03:41
  • 2
    Adding on to @WillJagy's comment, you can cut your work in half by noting $(-x)^3=-x^3$ so you only have to check the residues up to $6$. The others will give you the negatives of what you get from $0,\ldots,6$. – Cheerful Parsnip Sep 19 '23 at 03:43
  • 1
    V.E. it really would be a good thing if you found the cubes mod 13. For example, $ 3^3 = 27 = 26 + 1 \equiv 1 \pmod {13}.$ We can reduce at each stage: $5^2 = 25 \equiv -1 \pmod {13}$ so that $5^3 \equiv -1 \cdot 5 = -5 \equiv 8 \pmod {13}$ – Will Jagy Sep 19 '23 at 03:49
  • Applying Euler's Criterion as here in the dupe: $\bmod 13!:\ \left[a^3\equiv 3\right ]^4\Rightarrow 1\equiv 3\Rightarrow 13\mid 2\Rightarrow!\Leftarrow$ – Bill Dubuque Sep 19 '23 at 04:22

2 Answers2

5

Here's another, I believe easier, way. We have

$$((x-2)^3)^4 \equiv 3^4 \pmod{13} \;\;\to\;\; (x-2)^{12} \equiv 81 \equiv 3 \pmod{13}$$

However, since $x-2 \not\equiv 0\pmod{13}$, then Fermat's little theorem gives that $(x-2)^{12} \equiv 1 \pmod{13}$, contradicting the above equation.

An alternate, although similar, method is to note that $(x-2)^6 \equiv 3^2 \equiv 9 \pmod{13}$, but $(x-2)^6 \equiv \pm 1 \pmod{13}$ (using that $a^{\frac{p-1}{2}}\equiv \pm 1\pmod{p}$ for $a \not\equiv 0\pmod{p}$), so once again we get a contradiction.

John Omielan
  • 47,976
0

We can ask the simpler question does there exist an $x$ such that $x^3\equiv 3 \pmod {13}.$ (expecting that no such number will exist.

$2$ generates the multiplicative group $\mathbb Z_{13}^\times$... that is all numbers (other than zero) can be represented as $2^k$

But $2^{3k} \equiv \{8,12,5,1\}$ and $3$ is not in that set.

user317176
  • 11,017